I am trying to learn more about probability and came across an interesting question that I am stuck on and can no longer find online. There are 20 numbered balls and 10 bins. Someone is trying to assign the balls to the bins, but does it with replacement on accident.
So they did the following: Place a ball in bin 1, record it, then remove ball (with replacement remember). Place a ball in bin 2, record it, then remove ball. Place a ball in bin 3, record it, then remove ball. So for each bin, you have put in 1 ball. There are ten bins, therefore you do that process once for every bin. Once you have done that the experiment is over.
What is the probability exactly 1 ball was assigned to exactly 4 bins? What is the probability at least 2 bins received the same ball?
A) 1 Ball in 4 Bins:
We have ${20 \choose 1}$ being the different ways we can choose the 1 ball that was assigned. Also, we have ${19 \choose 6}$ being the different ways the other 19 balls can be picked for assignment. However, what is the sample size? Would it be $20^{10}$? Thus the answer would be $\frac{{20 \choose 1}{19 \choose 6}}{20^{10}}$.
B) Probability of at least 2 repeated can be represented as $1-P(\text{Zero Repeated})- P(\text{One Repeated})$. So $P(0) = {20 \choose 10}/20^{10}$ and $P(1) = \frac{{20 \choose 1}{19 \choose 9}}{20^{10}}$. Then we can plug and chug.
Are these right? Is this how to think about this type of problem?
Best Answer
After clarification of the question it can be answered as following:
We have $\binom{20}{1}$ ways to choose the "4-fold" ball and $\binom{10}{4}$ ways to choose the bins where it should go. The other 6 bins can be arbitrarily filled with remaining 19 balls. Hence the overall number of combinations is: $$ \binom{20}{1}\binom{10}{4}19^6. $$ In this way we however double-count the cases where there are two balls each assigned to exactly 4 bins. There are $\binom{20}{2}$ such pairs and $\binom{10}4\binom{6}4$ ways to choose corresponding bins. The other 2 bins can be filled arbitrarily with remaining 18 balls. Bringing everything together the final result is: $$ \frac{\binom{20}{1}\binom{10}{4}19^6-\binom{20}{2}\binom{10}{4}\binom{6}{4}18^2}{20^{10}}. $$
The simplest way to answer this question is to use complementary probability of the event "all bins receive different balls": $$ 1-\frac{\frac{20!}{10!}}{20^{10}}. $$ Replacement of $\frac{20!}{10!}$ with $\binom{20}{10}$ would be wrong here, since after choosing $10$ balls out of $20$ there are still $10!$ ways to assign the balls to certain bins.